Question

Let $f\left(x\right)=2x+1$. Then the number of real values of $x$ for which the three unequal numbers $f\left(x\right),f\left(2x\right),f\left(4x\right)$ are in $GP$ is

• A. 1
• B. 2
• C. 0
• D. none of these

Answer 1

The correct option is C 0

Given function is,

$f\left(x\right)=2x+1$

$\Rightarrow f\left(2x\right)=2\left(2x\right)+1=4x+1$ and

$f\left(4x\right)=2\left(4x\right)+1=8x+1$

Now,

$f\left(x\right),f\left(2x\right),f\left(4x\right)$ are in G.P.

$\therefore f^2\left(2x\right)=f\left(x\right)\cdot f\left(4x\right)\Rightarrow (4x+1{)}^2=(2x+1\left)\right(8x+1)\Rightarrow 16x^2+8x+1=16x^2+10x+1\Rightarrow 2x=0\Rightarrow x=0$

But, for $x=0$,

$f\left(x\right)=f\left(2x\right)=f\left(4x\right)=1$

Hence, for no real values of $x$, the given unequal numbers $f\left(x\right),f\left(2x\right),f\left(4x\right)$ are in G.P.